11 THE SPECTRAL THEORY OF HUBERT 409 eigenvector of £/, then (/®/)~ is an eigenvector of the unitary operator corresponding to u x w, for the eigenvalue 1. To prove that (y) implies (a), use the last criterion of (b); if (/|T) = 0, introduce the measure v^m~ftj (15.11.1), and observe that this measure on U is diffuse (15.11.6). Then we are reduced to proving that u 1 dv(Q = 0. Write this relation in the form lim JJ (~J and remark that the diagonal of U x U is (v ® v)-negligible.) (e) With the notation of Section 13.9, Problem 13(c), show that if 6 is irrational, the mapping z\-+e**i9z of U onto U is not weakly mixing. (Calculate the spectrum of the corresponding unitary operator £/.) (f) Suppose that the space L£(X, p.) is the Hilbert sum of the subspace C • 1 (the classes of the constant functions) and an at most denumerable family (H/X/6j of Hilbert spaces, where each H, has a Hilbert basis (enJ\eZ such that U-enJ*= en + ! , j for all n e Z (cf . Problem 1 5). Then the mapping u is mixing. In particular, if. (X, JLA, p) is the Bernoulli scheme B(i, J) (Section 13.21, Problem 18) then u is mixing. (If, for each neZ,fn is the function on lz such that fn(x) « — 1 if prnx = 0, and /„(*) = 1 if prn x = 1, then the classes of the finite products /ni/«2 • • •/„*, in which all the indices are distinct, form together with the class of 1 an orthonormal basis of Lc(X, /A) Likewise show that, if X is the torus T2, TT : R->T the canonical homo- morphism and /x the normalized Haar measure on X, then u defined by \ rr(x + 2y)) is mixing. (g) Suppose that u is ergodic with respect to /x, so that (by virtue of (c)) the eigen- values of U form an at most denumerable subgroup G of U, the eigenspace correspon- ding to an eigenvalue a e G being a line D(a) in L£(X, /A). Show that there exists a family of eigenvectors A e D(a) of U such that |/« | = 1 almost everywhere in X and such that/a/, = /a/0 almost everywhere, for all pairs (a, ]3) of points of G. (Let ha e D(a) be such that |/z«| = 1 almost everywhere, for all aeG; for each pair of points a, |3 e G we may write ha0 = r(oc, /3)hah0 almost everywhere, where r(a, fi) e U is a constant. Denote by A the subgroup of Ux generated by the ha for a e G and the group of constant functions from X to U (which may be identified with U). Show that there exists a homomorphism 6 : A-*U such that 0(0 = £ fc>r all £ eU. For this purpose, arrange the h* in a sequence (hn) and proceed by induction: if 0 is already defined on the subgroup A« generated by U and the hj such thaty"</z, distinguish two cases according as h* is not constant almost everywhere for any nonzero k e Z, or on the contrary that there exists a smallest positive integer k such that /zf is constant almost everywhere; use the fact that for all j e U and all integers k > 0, there exists T] e U such that yk = £. Then take/« = 0(/za)/za.) (h) Let v : X -> X be another mapping which is ergodic with respect to /x, and let V be the corresponding unitary operator. Show that if u satisfies the hypotheses of (g) and if U, V have the same eigenvalues, then u and t; are conjugate (Section 13.12, Problem 11).= 1 for all AeK. Then there exists no operator T'e &(E)